Array initialization in Go - Mini Project: Build & Apply

Array initialization

📖 Scenario: You are working on a simple inventory system for a small store. You need to keep track of the quantity of five different products.

🎯 Goal: Create an array to store the quantities of five products, set a threshold for low stock, find which products are below this threshold, and print their quantities.

📋 What You'll Learn

Create an array called quantities with exactly five integer values: 10, 3, 7, 2, 5

Create an integer variable called lowStockThreshold and set it to 4

Use a for loop with index i to check each quantity in quantities

Create a new array called lowStock to store quantities below the threshold

Print the lowStock array

💡 Why This Matters

🌍 Real World

Stores and warehouses often track product quantities to know when to reorder items.

💼 Career

Understanding arrays and loops is essential for managing collections of data in software development.

1

Create the quantities array

Create an array called quantities with these exact integer values: 10, 3, 7, 2, 5.

Go

# Create the quantities array with 5 integers
# Your code here

Hint

Use Go's array syntax: varName := [size]type{values}

2

Set the low stock threshold

Create an integer variable called lowStockThreshold and set it to 4.

Go

package main

func main() {
    quantities := [5]int{10, 3, 7, 2, 5}
    // Create lowStockThreshold variable and set it to 4
    // Your code here
}

Hint

Use variableName := value to create and set a variable.

3

Find quantities below the threshold

Use a for loop with index i to check each quantity in quantities. Create a slice called lowStock and append quantities less than lowStockThreshold to it.

Go

package main

func main() {
    quantities := [5]int{10, 3, 7, 2, 5}
    lowStockThreshold := 4
    lowStock := []int{}
    // Use a for loop with i to check quantities and append low stock items
    // Your code here
}

Hint

Use len(array) to get the array length and append(slice, value) to add items to a slice.

4

Print the low stock quantities

Write a fmt.Println statement to print the lowStock slice.

Go

package main

import "fmt"

func main() {
    quantities := [5]int{10, 3, 7, 2, 5}
    lowStockThreshold := 4
    lowStock := []int{}
    for i := 0; i < len(quantities); i++ {
        if quantities[i] < lowStockThreshold {
            lowStock = append(lowStock, quantities[i])
        }
    }
    // Print the lowStock slice
    // Your code here
}

Hint

Use fmt.Println(lowStock) to display the slice.

Practice

1. Which of the following correctly declares an array of 5 integers in Go?

easy

A. var arr int[5]

B. var arr [5]int

C. arr := []int{5}

D. arr := [5]int{1, 2, 3, 4}

Solution

1. Step 1: Understand Go array declaration syntax

   In Go, arrays are declared with the syntax: var name [size]type.

2. Step 2: Check each option

   A: var arr int[5] wrong syntax (type before size). B: arr := []int{5} is a slice, not an array. C: var arr [5]int correct syntax. D: arr := [5]int{1, 2, 3, 4} too few initializers (4 < 5), compilation error.

3. Final Answer:

   var arr [5]int -> Option B

4. Quick Check:

   Array declaration = var arr [5]int [OK]

Hint: Remember: var name [size]type declares an array [OK]

Common Mistakes:

• Confusing slices with arrays
• Using wrong syntax like int[5]
• Missing var keyword for array declaration

2. Which of these is the correct way to initialize an array of 3 strings with values "a", "b", "c" in Go?

easy

A. arr := [3]string{"a", "b", "c"}

B. arr := []string{"a", "b", "c"}

C. var arr [3]string = {"a", "b", "c"}

D. arr := [3]string{"a", "b", c}

Solution

1. Step 1: Identify correct array initialization syntax

   Arrays use [size]type{values} syntax. arr := [3]string{"a", "b", "c"} matches this exactly.

2. Step 2: Check other options for errors

   arr := []string{"a", "b", "c"} is a slice, not an array. var arr [3]string = {"a", "b", "c"} uses invalid initialization with = and braces. arr := [3]string{"a", "b", c} misses quotes around "c" causing syntax error.

3. Final Answer:

   arr := [3]string{"a", "b", "c"} -> Option A

4. Quick Check:

   Array init with values = [3]string{"a", "b", "c"} [OK]

Hint: Use [size]type{values} for array initialization [OK]

Common Mistakes:

• Using slice syntax instead of array
• Missing quotes around string elements
• Incorrect use of = with array literals

3. What will be the output of this Go code?

package main
import "fmt"
func main() {
  arr := [4]int{1, 2}
  fmt.Println(arr)
}

medium

A. [1 2 0 0]

B. [1 2]

C. [0 0 0 0]

D. Compilation error

Solution

1. Step 1: Understand partial array initialization

   When fewer elements are provided than the array size, Go fills remaining elements with zero values.

2. Step 2: Apply to given array

   Array size is 4, initialized with 1 and 2. Remaining two elements become 0 by default.

3. Final Answer:

   [1 2 0 0] -> Option A

4. Quick Check:

   Partial init fills rest with zero [OK]

Hint: Uninitialized array elements default to zero [OK]

Common Mistakes:

• Assuming array length matches initializer length
• Expecting error for fewer initial values
• Confusing slices with arrays

4. Identify the error in this Go array initialization:

var arr [3]int = [3]int{1, 2, 3, 4}

medium

A. No error, code is correct

B. Missing var keyword

C. Incorrect type declaration

D. Array size mismatch with initializer

Solution

1. Step 1: Check array size and initializer length

   Array declared with size 3 but initializer has 4 elements.

2. Step 2: Understand Go array rules

   Go requires initializer length to match array size exactly, else compilation error occurs.

3. Final Answer:

   Array size mismatch with initializer -> Option D

4. Quick Check:

   Initializer length must equal array size [OK]

Hint: Initializer length must match array size exactly [OK]

Common Mistakes:

• Ignoring extra elements in initializer
• Assuming Go auto-expands array size
• Confusing slices with arrays

5. You want to create an array of 5 integers where the first two elements are 10 and 20, and the rest are zeros. Which code snippet correctly achieves this and prints the array?

hard

A. arr := [5]int{10, 20, 30}
fmt.Println(arr)

B. arr := [3]int{10, 20}
fmt.Println(arr)

C. arr := [5]int{10, 20, 0, 0, 0}
fmt.Println(arr)

D. arr := []int{10, 20}
fmt.Println(arr)

Solution

1. Step 1: Understand array initialization with explicit zeros

   To ensure the last three elements are zeros explicitly, include them in the initializer.

2. Step 2: Check each option

   B: arr := [3]int{10, 20} fmt.Println(arr) wrong size 3, not 5. A: arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) explicitly sets zeros. C: arr := [5]int{10, 20, 30} fmt.Println(arr) third element 30, not zero. D: arr := []int{10, 20} fmt.Println(arr) is a slice, not array.

3. Step 3: Choose the best explicit solution

   A clearly shows all elements matching the requirement exactly.

4. Final Answer:

   arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) -> Option C

5. Quick Check:

   Explicit zeros in initializer = arr := [5]int{10, 20, 0, 0, 0} fmt.Println(arr) [OK]

Hint: Explicitly list zeros to avoid confusion [OK]

Common Mistakes:

• Assuming missing elements are always zero without clarity
• Using slices instead of arrays
• Adding unintended values in initializer